Is $A$ full or sparse? Are you doing this once or many times with the same $A$?
–
jorikiApr 1 '11 at 16:56

A is not sparse and has many nonzero elements, however, the coefficients themselves are derived from a smaller set of variables.
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mghandiApr 2 '11 at 3:37

About your second question, yes I'm doing this many times. I am familiar with the LU decomposition trick to speed up when the coeffs matrix is unchanged. Is there any other tricks to do that even more efficiently?
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mghandiApr 2 '11 at 3:51

1

You may apply iterative methods (CG, if you matrix is spd, GMRES something similar otherwise). You may also want to ask at scicomp.stackexchange.com.
–
DirkDec 15 '12 at 17:43

3 Answers
3

Unless your matrix is sparse or structured (e.g. Vandermonde, Hankel, or those other named matrix families that admit a fast solution method), there is not much hope of doing things better than $O(n^3)$ effort. Even if one were to restrict himself to solving for just one of the 50,000 variables, Cramer will demand computing two determinants for your answer, and the effort for computing a determinant is at least as much as decomposing/inverting a matrix to begin with.

There is a way to reduce the complexity and make the system solvable in parallel.
It is called Diakoptics (a method invented by Gabriel Kron). The methods primary use is for large electrical networks that have few interconnections like power grids. But you should be able to adapt it.

The complexity (for the case below) is reduced from $O(n^3)$ to $O(2(\frac{n}{2})^3)$ or $O(\frac{1}{4}n^3)$, the impact can be much greater if the system is divided it into more subsystems. For that case the complexity is ($s$-subsysems, $c$-interconnection points) $O(c^3)+O((\frac{n^3}{s²}))$, if the systems is divided into equaly sized subsystems. I'm not sure about the notation for multiple variables, but you should get the point.

In short:

Lets assume you have a $N \times N$ system, lets say you can divide the system into two systems with 1 connection point(plus reference when you look at electrical systems). The connection points are $m$ and $n$. Lets assume these systems are of the size $N_1=N/2$ and $N_2=N/2$ (for simplicitys sake). You should now solve them separately.

$\mathbf A_1^{-1}=\mathbf B_1$

$\mathbf A_2^{-1}=\mathbf B_2$

The next step is to put them back together, that is done with the help of the so called "Thevenin Matrix"(in our case it is 1$\times$1). You can look up the exact principle for higher orders(more connection points), but for this example it looks like:
\begin{align}
\mathbf{B_{TH}}=B_{1mm}+B_{2nn}-2B_{mn}
\end{align}
For our case we have $B_{mn}=0$. Now we need the solutions $x_1$ and $x_2$ to form the coefficients $b_{th}$.

The $\mathbf b_{th}$ matrix only has nonzero elements at $m$ an $N/2 +n$. Now we can finally find the solution $x_n$ for the whole system:
\begin{align}
\mathbf x_n=\begin{bmatrix}x_1\\x_2
\end{bmatrix}-\begin{bmatrix}B_1&0\\0&B_2
\end{bmatrix}\begin{bmatrix}b_{th}
\end{bmatrix}
\end{align}

I'm more used to the engineering notation with $Z, I, U$ and so on, so excuse for non-standard symbol usage.

[this should be a comment to @Juan Joder's answer or to the OP's question, but not enough repo]

strang gilbert - linear algebra and its applications 3rd edition page 16 at the footer mentions that complexity had fallen already fallen below $O(Cn^{2.376})$ at the date of writing 1988, altough $C$ is so large that makes the algorithm impractical for most matrix sizes found in practice today. It does not mention the name of the algorithm though.

personal guess: not sure about solving for single variables, but I don't think you can reduce complexity like that since the entire system is coupled.